Question

Let $z_{k}(k=0,1,2, \ldots, 6)$ be the roots of the equation $(z+1)^{7}+z^{7}=0$, then $\displaystyle\sum_{k=0}^{6} \text{Re}\left(z_{k}\right)$ is equal to

• A. $0$
• B. $\frac{3}{2}$
• C. $-\frac{7}{2}$
• D. $\frac{7}{2}$

Answer 1

Answer: (C)

we have $\left(z_{k}+1\right)^{7}+z_{k}^{7}=0$
$\Rightarrow \left(z_{k}+1\right)^{7}=-z_{k}^{7} $
$\Rightarrow \left|z_{k}+1\right|^{7}=\left|z_{k}\right|^{7}$
$\Rightarrow \left|z_{k}+1\right|=\left|z_{k}\right| $
$\Rightarrow \left|x_{k}+i y_{k}+1\right|^{2}=\left|x_{k}+i y_{k}\right|^{2}$
$\Rightarrow \left(x_{k}+1\right)^{2}+y_{k}^{2}=x_{k}^{2}+y_{k}^{2}$
$\Rightarrow 2 x_{k}+1=0 $
or $x_{k}=-\frac{1}{2}$
Thus, $\displaystyle\sum_{k=0}^{6} \text{Re}\left(z_{k}\right)=\displaystyle\sum_{k=0}^{6} x_{k}=-\frac{7}{2}$.